Question

Two particles are projected from the same point with the same speed $u$ such that they have the same range $R$, but different maximum heights ${h_1}$ and ${h_2}$. Which of the following is correct?
(A) ${R^2} = 2{h_2}{h_2}$
(B) ${R^2} = 16{h_1}{h_2}$
(C) ${R^2} = 4{h_1}{h_2}$
(D) ${R^2} = {h_1}{h_2}$

Answer 1

In this question two particles have been projected from same point with the same speed $u$ such that, they have same range $R$ , therefore, for the same range angle of projection will be $\theta$ and $90 - \theta$, that is ,complimentary. Hence by using the concept of projectile motion we will solve this question.

Complete step by step solution:
According to the question, particles are projected from same point with the same speed $u$ such that, they have same range $R$ ,
We know that $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
$\Rightarrow R = \dfrac{{{u^2}2\sin \theta \cos \theta }}{g}$
Hence, for angle $\theta$ ,
${h_1} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} - - - (1)$
And For angle $90 - \theta$ ,
${h_2} = \dfrac{{{u^2}{{\cos }^2}\theta }}{{2g}} - - - (2)$
Now, multiplying equation $(1)$ and $(2)$ we have,
${h_1}{h_2} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \times \dfrac{{{u^2}{{\cos }^2}\theta }}{{2g}}$
$\Rightarrow {\left[ {\dfrac{{{u^2}\sin \theta \cos \theta }}{{2g}}} \right]^2} \times \dfrac{4}{4}$
$\Rightarrow \dfrac{1}{{16}}{\left[ {\dfrac{{{u^2}\sin 2\theta }}{g}} \right]^2}$
$\Rightarrow {h_1}{h_2} = \dfrac{1}{{16}}{R^2} \\\ \Rightarrow {R^2} = 16{h_1}{h_2} \\\$
Therefore, option (B) is the correct answer.

Note:
We must keep in mind that if two particles projected with the same range then the angle they will have between them will be complementary, that is $\theta$ and $90 - \theta$ .